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3 years ago

10

Freddie drives his car 10 km North. He stops for gas and then drives 5 km South.

1 answer:

fgiga [73]3 years ago

4 0

Answer:

5 km

Explanation:

displacement is the straight line distance from a particular point. since north and south are in opposite directions, it means that he has taken a u-turn

thus, 10 - 5 = 5km displacement

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Which of the following codes is used to report the removal of 37 skin tags by electrosurgical destruction?

Tema [17]

Are there any answer choices??

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3 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

a speed swimmer love to race around the parks pine which is 25m around if she can swim 20 laps in 7200s what is her average spee

aleksley [76]

Answer:

the average speed of the swimmer is 0.069 m/s.

Explanation:

Given;

complete distance around the park pine, d = 25 m

total lap completed, = 20 laps

time of laps completion, t = 7200 s

The total distance completed by the swimmer = 20 x 25 = 500 m

The average speed of the swimmer = distance / time

= (500 m) / (7200 s)

= 0.069 m/s.

Therefore, the average speed of the swimmer is 0.069 m/s.

8 0

3 years ago

A baseball with a mass of 300 g has a kinetic energy of 304 J. Calculate the speed of the baseball

scZoUnD [109]

The formula for kinetic energy is equal to 1/2mv^2, where "m" is the mass of the object (in kilograms) and "v" is equal to the velocity of the object (in meters per second). To calculate the speed, simply plug in the values and solve.

KE = 0.5mv^2

304 J = 0.5(0.3 kg)v^2 -mass converted from grams to kilograms

v = 45.02 m/s

The baseball is travelling about 45.02 meters per second.

Hope this helps!

4 0

3 years ago

Read 2 more answers

A mass of 150 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity

nadezda [96]

Answer:

$u(t)=\frac{1}{5} sin\ (25t)$

Explanation:

Given:

mass of the body stretching the spring, $m=150\ g$

extension in spring, $\Delta x=1.568\ cm$

velocity of oscillation, $u'(0)=20\ cm.s^{-1}$

initial displacement position of equilibrium, $u(0)=0$

<u>According to given:</u>

$m.g=k.\Delta x$

$150\times 980=k\times 1.568$

$k=93750\ dyne.cm^{-1}$

<u>we know frequency:</u>

$\omega=\sqrt{\frac{k}{m} }$

$\omega=\sqrt{\frac{93750}{150} }$

$\omega=25$

Now, for position of mass in oscillation:

$u= A.sin\ (\omega.t)+B.cos\ (\omega.t)$

$u= A.sin\ (25.t)+B.cos\ (25.t)$

at $t=0;\ u(0)=0\ \Rightarrow A=0$

∴ $u(t)=B.sin\ (25.t)$

∵ at $t=0;\ u'(0)=20\ cm.s^{-1}\ \Rightarrow B=\frac{1}{5}$

$u(t)=\frac{1}{5} sin\ (25t)$

7 0

3 years ago